 by Daniel Phillips

### Categories

Codility has a number of lessons online to help candidates prepare for the problems on the site. I figured it might be worthwhile to make a summary of some of the algorithms from the lessons that I more easily forget.

# Prefix (and suffix) sums

Makes it easy to calculate totals for slices of an array. Codility provides the following algorithm:

def prefix_sums(A):
n = len(A)
P =  * (n + 1)
for k in range(1, n + 1):
P[k] = P[k - 1] + A[k - 1]
return P


Note that the first element is 0 with this algorithm.

The sum of the elements in a slice from x to y can be calculated using P[y + 1] - P[x].

The leader of an array is the element that occurs more than $$\frac{n}{2}$$ times in the array. Codility provides the following $$O(n)$$ solution for this:

def goldenLeader(A):
n = len(A)
size = 0
# keep a 'stack' of leaders to find a potential candidate
for k in xrange(n):
if (size == 0):
size += 1
value = A[k]
else:
if (value != A[k]):
size -= 1
else:
size += 1

# validate that the candidate is actually the leader
candidate = -1
if (size > 0):
candidate = value
count = 0
for k in range(n):
if (A[k] == candidate):
count += 1

if(count > n // 2):

# returns -1 if no leader was found


# Maximum slice

Finds the slice with the largest sum. Codility provides an implementation of Kadane’s algorithm for this problem:

def golden_max_slice(A):
max_ending = max_slice = 0
for a in A:
max_ending = max(0, max_ending + a)
max_slice = max(max_slice, max_ending)
return max_slice


Initialisation doesn’t have to start from 0 and Wikipedia has more information on variations of the algorithm.

# Counting divisors of $$n$$

Codility provides an algorithm that can count divisors of $$n$$ in $$O(\sqrt{n})$$. It relies on finding symmetric divisors, which allows you to count two divisors for the price of one.

def divisors(n):
i = 1
result = 0
# iterate up to sqrt(n)
while (i * i < n):
if (n % i == 0):
result += 2
i += 1
# if i^2 == n, then we count an extra divisor
if (i * i == n):
result += 1
return result


## Checking for prime numbers

Based on the same principles as the example above, you can check for primes by iterating over i * i <= n. If n % i == 0, then the number isn’t prime.

## Sieve of Eratosthenes

Can be used to find all prime numbers in the range $$2$$ to $$n$$.

Here’s an implementation based on Codility’s notes.

Some quick reminders:

• Need a list of flags for each number from 2 to n.
• Main loop iterates from 2 to $$\sqrt{n}$$ (inclusive). Not necessary to iterate all the way to n.
• Each time the main loop is incremented, if the flag for the current value i indicates it is still prime:
1. Mark $$i^2$$ as non prime.
2. Mark $$i^2 + i$$ as non prime.
3. Continue marking multiples as non prime until reaching n.
• At the end, all prime numbers will have been found.
• Complexity is $$O(n\log{\log{n}})$$.

### Finding prime factors

The Sieve of Eratosthenes algorithm can be modified to find the prime factors of a number n.

1. Build an array F of minimum prime factor for each value of i instead of just using true/false.
2. Iterate through this array starting at element n.
3. Store the prime factor at position F[n] in the output list of prime factors.
4. n = n / F[n]
5. Keep iterating until F[n] == 0

Complexity is $$O(\log{n})$$.

See here for an implementation based on Codility’s notes.

### Euclidean algorithm

The Codility notes provide 3 approaches for finding the greatest common divisor (gcd) between 2 numbers:

• Euclidean algorithm by subtraction: recursively subtract the larger value from the smaller until the values are equal. $$O(n)$$
• Euclidean algorithm by division: recursively use the modulo operator until the two values are divisible by each other. $$O(\log(a + b))$$ where $$a$$ and $$b$$ are the two input values.
• Binary Euclidean algorithm: a more complex implementation involving division by 2 and scaling the result. $$O(\log{n})$$

The least common multiple (lcm) of $$a$$ and $$b$$ is the smallest value that can be divided by $$a$$ and $$b$$.

$lcm(a, b) = \frac{a \cdot b}{gcd(a, b)}$

See here for an implementation based on Codility’s notes.

### Fibonacci numbers

def fib(n):
fib =  * (n + 2)
fib = 1

for i in range(2, n + 1):
fib[i] = fib[n - 1] + fib[n - 2]

return fib[n]


Complexity is $$O(n)$$.

Start with a sorted list of values v, length n. beg = 0, end = n - 1, $${ mid = \lfloor\frac{beg + end}{2}}\rfloor$$. Compare the value of v[mid] to x, and set beg or end to mid ± 1 depending on the value. Keep repeating until x has been found (or can’t be found).

Complexity is $$O(\log{n})$$.